Calculate the energy of the photon emitted when an electron in a He⁺ ion transitions from the second excited state to the first excited state. (Given: Ionization energy of H atom in the ground state = 2.18 10⁻¹⁸ J atom ⁻¹ )

Step-by-step solution

For a hydrogen-like species, the energy difference between two states is given by: E = IE_ H Z^2 ( 1 n₁^2 - 1 n₂^2 ) For the He⁺ ion, the atomic number Z = 2 . The first excited state corresponds to n₁ = 2 . The second excited state corresponds to n₂ = 3 . Substituting the given values into the formula: E = 2.18 10⁻¹⁸ (2)^2 ( 1 2^2 - 1 3^2 ) E = 2.18 10⁻¹⁸ 4 ( 1 4 - 1 9 ) E = 2.18 10⁻¹⁸ 4 ( 9 - 4 36 ) E = 2.18 10⁻¹⁸ 4 5 36 E = 2.18 10⁻¹⁸ 5 9 E 1.21 10⁻¹⁸ J Answer: 1.21 10⁻¹⁸ J